Equivalent air depth

The equivalent air depth can be calculated for depths in metres as follows:

EAD = (Depth + 10) × Fraction of N₂ / 0.79 − 10

Working the earlier example, for a nitrox mix containing 64% nitrogen (EAN36) being used at 27 metres, the EAD is:

EAD = (27 + 10) × 0.64 / 0.79 − 10

EAD = 37 × 0.81 − 10

EAD = 30 − 10

EAD = 20 metres

So at 27 metres on this mix, the diver would calculate their decompression requirements as if on air at 20 metres.

The equivalent air depth can be calculated for depths in feet as follows:

EAD = (Depth + 33) × Fraction of N₂ / 0.79 − 33

Working the earlier example, for a nitrox mix containing 64% nitrogen (EAN36) being used at 90 feet, the EAD is:

EAD = (90 + 33) × 0.64 / 0.79 − 33

EAD = 123 × 0.81 − 33

EAD = 100 − 33

EAD = 67 feet

So at 90 feet on this mix, the diver would calculate their decompression requirements as if on air at 67 feet.

For a given nitrox mixture and a given depth, the equivalent air depth expresses the theoretical depth that would produce the same partial pressure of nitrogen if regular air (79% nitrogen) was used instead:

${\displaystyle ppN_{2}(nitrox,depth)=ppN_{2}(air,EAD)}$

Hence, following the definition of partial pressure:

${\displaystyle FN_{2}(nitrox)\cdot P_{depth}=FN_{2}(air)\cdot P_{EAD}}$

with ${\displaystyle FN_{2}}$ expressing the fraction of nitrogen and ${\displaystyle P_{depth}}$ expressing the pressure at the given depth. Solving for ${\displaystyle P_{EAD}}$ then yields a general formula:

${\displaystyle P_{EAD}={FN_{2}(nitrox) \over FN_{2}(air)}\cdot P_{depth}}$

In this formula, ${\displaystyle P_{EAD}\,}$ and ${\displaystyle P_{depth}\,}$ are absolute pressures. In practice, it is much more convenient to work with the equivalent columns of seawater depth, because the depth can be read off directly from the depth gauge or dive computer. The relationship between pressure and depth is governed by Pascal's law:

${\displaystyle P_{depth}=P_{atmosphere}+\rho _{seawater}\cdot g\cdot h_{depth}\,}$

Using the SI system with pressures expressed in pascal, we have:

${\displaystyle P_{depth}(Pa)=P_{atmosphere}(Pa)+\rho _{seawater}\cdot g\cdot h_{depth}(m)\,}$

Expressing the pressures in atmospheres yields a convenient formula (1 atm ≡ 101325 Pa):

${\displaystyle P_{depth}(atm)=1+{\frac {\rho _{seawater}\cdot g\cdot h_{depth}}{P_{atmosphere}(Pa)}}=1+{\frac {1027\cdot 9.8\cdot h_{depth}}{101325}}\ \approx 1+{\frac {h_{depth}(m)}{10}}}$

To simplify the algebra we will define ${\displaystyle {\frac {FN_{2}(nitrox)}{FN_{2}(air)}}=R}$. Combining the general formula and Pascal's law, we have:

${\displaystyle 1+{\frac {h_{EAD}}{10}}=R\cdot (1+{\frac {h_{depth}}{10}})}$

so that

${\displaystyle h_{EAD}=10\cdot (R+R\cdot {\frac {h_{depth}}{10}}-1)=R\cdot (h_{depth}+10)-10}$

Since ${\displaystyle h(ft)\approx 3.3\cdot h(m)\,}$, the equivalent formula for the imperial system becomes

${\displaystyle h_{EAD}(ft)=3.3\cdot {\Bigl (}R\cdot ({\frac {h_{depth}(ft)}{3.3}}+10)-10{\Bigr )}=R\cdot (h_{depth}(ft)+33)-33}$

Substituting R again, and noting that ${\displaystyle FN_{2}(air)=0.79}$, we have the concrete formulas:

${\displaystyle h_{EAD}(m)={\frac {FN_{2}(nitrox)}{0.79}}\cdot (h_{depth}(m)+10)-10}$

${\displaystyle h_{EAD}(ft)={\frac {FN_{2}(nitrox)}{0.79}}\cdot (h_{depth}(ft)+33)-33}$

Although not all dive tables are recommended for use in this way, the Bühlmann tables are suitable for use with these kind of calculations. At 27 metres the Bühlmann 1986 table (0–700 m) allows 20 minutes bottom time without requiring a decompression stop. While at 20 metres the no-stop time is 35 minutes. This shows that using EAN36 for a 27-metre dive can give a 75% increase in bottom time over using air.

US Navy tables have also been used with equivalent air depth, with similar effect. The calculations are theoretically valid for all Haldanean decompression models.